Question

Newton's law of cooling is T = A * e ^ (- d * t) + C where is the temperature of the object at time and C is the constant temperature of the surrounding mediumSuppose that the room temperature is 71^ + and the temperature of a cup of tea 160when it is placed on the table. How long will it take for the tea to cool to 120 degrees for k = 0.0595943 Round your answer to two decimal places.

Answer 1

Given

`\begin{gathered} T=Ae^(-kt)+C\text{ --------\lparen1\rparen} \\ \\ C=71 \\ \\ A=160-71 \\ \\ T=120 \\ \\ k=0.0595943 \end{gathered}`

To find the time, we nee to substitute the C, A, T, and k in (1) and then determine (t

`\begin{gathered} 120=(160-71)e^(-0.0595943t)+71 \\ \\ \Rightarrow(120-71)/(160-71)=e^(-0.0595943t) \\ \\ \Rightarrow(49)/(89)=e^(-0.0595943t) \\ \\ \Rightarrow-0.0595943t=\ln((49)/(89)) \\ \\ \Rightarrow t=(1)/(-0.0595943)\ln((49)/(89))=10.01456\text{ s} \end{gathered}`

`t=(10.01465)/(60)\text{ mins}=0.17\text{ mins}`